I am given that X is a standard normal distribution.

Why is $Cov(X, -X) = -1$?

I know that $Cov(X,X) = Var(X)$ and that the $Var(X) = 1$.

Is the $Var(-X) = -1$?


This follows from $Cov(X,-X)=-Cov(X,X)$. In fact, more generally, $Cov(aX,bY)=abCov(X,Y)$.

Also, the above property can be used to show $Var(-X)=(-1)^2Var(X)=Var(X)$ so $Var(-X)\neq -1$ (in fact, the variance is always nonnegative).


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